$\dfrac{dy}{dt}=-7y$, and $y=3$ when $t=0$. Solve the equation. Choose 1 answer: Choose 1 answer: (Choice A) A $y=3+e^{-7t}$ (Choice B) B $y=3e^{-7t}$ (Choice C) C $y=e^{-7t}$ (Choice D) D $y=2+e^{-7t}$
The general solution of equations of the form $\dfrac{dy}{dt}=ky$ is $y=C\cdot e^{kt}$ for some constant $C$. This can be found using separation of variables. In our case, $k=-7$, so $y=C\cdot e^{-7t}$. Let's use the fact that $y=3$ when $t=0$ to find $C$ : $\begin{aligned} y&=C\cdot e^{-7t} \\\\ 3&=C\cdot e^{-7\cdot 0} \gray{\text{Plug }t=0\text{ and }y=3} \\\\ 3&=C \end{aligned}$ In conclusion, $y=3e^{-7t}$.